Sin6x cos6x 1 3 sin2x cos2x

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      = sin2x . cos2x - sin2x . cos2x 2(cos6x + sin6x) - 3(cos4x + sin4x) = 2(cos2x + + sin4x - sinx2 cos2x) - 3(cos4x + sin x) ... —-2 + 3 cos2x(1 + tanx) 1 + tan x + tanx 2+3 x 1 + tan x I + tanx 1 + tanx I + tan x 1 + tanx 1 + tan x 1 + tanx = 1+ tan2x tanx = tan2x tanx - tan2x = 0

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      = sin2x . cos2x - sin2x . cos2x 2(cos6x + sin6x) - 3(cos4x + sin4x) = 2(cos2x + + sin4x - sinx2 cos2x) - 3(cos4x + sin x) ... —-2 + 3 cos2x(1 + tanx) 1 + tan x + tanx 2+3 x 1 + tan x I + tanx 1 + tanx I + tan x 1 + tanx 1 + tan x 1 + tanx = 1+ tan2x tanx = tan2x tanx - tan2x = 0

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      FORMS OF INTEGRATION In order to evaluate this type of integration we have to use the following algorithm Step 1 – Open the polynomial by binomial Step 2 -Divide the polynomial and write it in mixed fraction Step 3 – put the value and integrate QUESTIONS

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      cos6x sin6x cos2x sin2x :هب تسا یواسم )1 2)2 1)3 2)4 :لح cos6x sin6x? cos2x sin2x cos6x sin6x cos6x sin2x sin6x cos2x sin 2x 6x cos2x sin2x cos2x sin2x cos2x sin2x 2sin4x 2sin4x 2sin2x cos2x sin4x 2:لح.16 رگا x 4 ; 1 x 22 fx 1 x 1 ; 0 x 1 2 d dd ­ ° ® ° ¯ سپ ،دشاب 3 f 2 §· c¨¸ ©¹:هب تسا یواسم )1 3)2 ...

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      The function defined on the set C* of all nonzero complex numbers satisfies the equation f(z)f(iz) = z2, for all z ∈ C∗. Prove that the function f(z) is odd, i,e., f(−z) = −f(z) for all z ∈ C∗. Give an example of a function that satisfies this condition. Solution. Note that f(z) 6= 0 for all x ∈ C∗.

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      3 1 sinx 3 1 cosx 1 3 . 10). 3sin3x 3cos9x 1 4sin 3x 3 LỜI GIẢI 1). 3 3cos2x cosx 1 2sinx . Điều kiện sinx 0 x k 1 3 3cos2x 2sinxcosx 3cos2x sin2x 3 3 1 3 cos2x sin2x 2 2 2 3 cos2x.cos sin2x.sin 6 6 2 2x k2 6 6 x k cos 2x cos , k6 6 6

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      sin4x —cos4x+sin2x cos2x , then sin4x+cos4x+sin2x cos2x (a)-3/2 /2 (c)-5/3 ... The maximum value of y= is sin6x+cos6x . The maximum value of cos2(450 + x)+ (sin x — cos x)2 is . If y = tan x/ tan , then find maximum and minimum value of . The maximum value of the expression sin 29 +3sin9cose +5cos2e . Created Date: 5/18/2022 6:42:14 AM

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      + cos2x = 1, the numerical value of (sin12x + 3sin10x + 3sin8x + sin6x – 1) is :Answer:Identity used:⇒ 1 – cos2x = sin2xGiven,→ cosx + cos2x = 1→ cosx = 1 – cos2x→ cosx = sin2xReplacing the value of sin2x as cosx in given expression:= sin12x + 3sin10x + 3sin8x + sin6x – 1= cos6x + 3cos5x + 3cos4x + cos3x – 1= cos6x + cos5x ...

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      2. sin6x + cos6x = 1 —3 sin2 x cos2 x Solution: Let us consider LHS: sin6x + cos6 x (sin2 x) 3 + (cos2 x) 3 ... 1 — (sin2x + cos2x) + sin x cos x We know, sin2x + cos2x — 1 —l + sin x cos x — sin X COS sin x cos2x = 2 + sin2 cos2 sin cos Sin x cos x RHS . LHS RHS Hence proved.

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      — sin6x + coss x4-3sin2x.cos2x; 27 27 + cos2 - 2sin2r = cos2 —X + COS2 — -Ex Ta có: f(x) = 2x 4 _ cos_ vay — eos£ 2 T biét — và (x) 4x b. Ðät u = 6 V — 7X—3 vs y' — (u.v)' — u'v + uv' + (7x-3)+7 c.y' (x —2)' x 2+1 x(x—2) 2x — 2x+l 2 tan x — (cot2x)' cos x sm x x x x sin sin

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      1e 2x cos6x+ c 2e 2x sin6x. Math 2250 Exam 2, Page 3 of 10 November 13, 2014 3. [8 pts.] ... 1 cos2x+ c 2 sin2x: (2) The annhilating operator for the RHS (right hand side) ... 1 3 5 (c) T F If the determinant of a matrix is 0 then that matrix is invertible. (d) T F A fourth order, linear, homogeneous DE with constant coe cients has ...

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      = sin2x . cos2x - sin2x . cos2x 2(cos6x + sin6x) - 3(cos4x + sin4x) = 2(cos2x + + sin4x - sinx2 cos2x) - 3(cos4x + sin x) ... —-2 + 3 cos2x(1 + tanx) 1 + tan x + tanx 2+3 x 1 + tan x I + tanx 1 + tanx I + tan x 1 + tanx 1 + tan x 1 + tanx = 1+ tan2x tanx = tan2x tanx - tan2x = 0

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      1 p 3 sin2x= sin 3x 2 cos x 2; á) Íàéäèòå êîðíè óðàâíåíèÿ sin6x sinx+ cosx = cos6x cosx sinx, äîâëåòâîðó ÿþùèå óñëîâèþ 0

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      using the standard trigonometric identities for sinasinb, cosacosb, cos2a and sin2a, we have that √ 3 = tan2x sin(2x+(π/3))sin(2x+(2π/3)) cos(2x+(π/3))cos(2x+(2π/3)) = tan2x cos(π/3)−cos(4x+π) cos(π/3)+cos(4x+π) = tan2x 1+2cos4x 1−2cos4x = tan2x 1+2(2cos22x−1) −2(1 2sin22x) = sin2x cos2x 4cos22x−1 4sin22x−1 = 2sin4xcos2x−sin2x …

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      1 2 cos2 2x = 1 4 (cos6x+ cos2x) + 1 4 (1 + cos4x) = 1 4 cos6x+ 1 4 cos4x+ 1 4 cos2x+ 1 4: ˇ, I= 1 24 sin6x+ 1 16 sin4x+ 1 8 sin2x+ 1 4 x+ C: 3/15 kJ Ik J I ‹£ ˝¶ ’4 òÑ ...

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      20. Z sin 10xsin 4x dx Solution: We wil use the product-to-sum identities to trasform this product into a sum. We write the cosineformula for the sum and the di¤erence of these two angles. cos 14xcos 6x= cos (10x+ 4x) = cos 10xcos 4xsin 10xsin 4x= cos (10x4x) = cos 10xcos 4x+ sin 10xsin 4x We will subtract the rst equation from the rst one

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      3 p x ( sin2x)(2)+(cos2x) 1 3 3 p x2 8. h0(x) = (x3 +6x 2)7(2)(9x+1)(9)+(9x+1)7(7)(x3 +6x2)6(3x +12x ... csc2 1 3 x 1 3 + cot 1 3 x 7 3 p x4 12. g0(x) = 3 2 p 2 cos6x (sin6x)(6) 13. dy dx = p x (cos2x)(2)+(sin2x) 1 2 p x 14. f0(x) = (x2 +2x)( csc3xcot3x)(3) (csc3x)(2x+2) (x 2+2x) 15. h0(x) = 2tan 1 3 x sec2 1 3 x 1 3 16. dy dx = ( 2csc 5x)(5) 3 ...

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      sin6x sin2x = lim x→0 (cos6x)·6 (cos2x)·2 . Since both cos6x and cos2x obviously approach 1 as x → 0, we get lim x→0 sin6x sin2x =6 2 = 3. Without L’Hˆopital’s Rule, lim x→0 sin6x sin2x = lim x→03 (sin6x)/(6x) (sin2x)(2x) = 3. 6. lim x→∞ (lnx)2 x Solution: UsingL’Hˆopital’sRule, lim x→∞ (lnx)2 x = lim x→∞ (2lnx)·1 x 1 = lim

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